Problem: What is the area of the region between the graphs of $f(x)=\sqrt{x+10}$ and $g(x)=x-2$ from $x=-10$ to $x=6$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{64}{3}$ (Choice B) B $\dfrac{320}{3}$ (Choice C) C $160$ (Choice D) D $128$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${2}$ ${4}$ ${6}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${\llap{-}10}$ ${2}$ ${4}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${\llap{-}10}$ ${\llap{-}12}$ ${\llap{-}14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=-10$ and $x=6$. From this we are looking to evaluate: $ \int_{-10}^{6}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-10}^{6} \left( \sqrt{x+10}- (x-2) \right) \,dx \\\\ &= \dfrac{2(x+10)^{3/2}}{3}-\dfrac{x^2}{2}+2x~\Bigg|_{-10}^{6} \\\\ &= \left( \dfrac{128}{3}-18+12 \right) - \left( 0-50-20 \right)\\\\ &= \dfrac{320}{3} \end{aligned}$ Answer The area is $\dfrac{320}{3}$ square units.